Question
Provide Write 2 the Question Content attribution L your answer overall Interpret Cell Notation to Describe _ (aq ) below: reaction that ~Caq) Rencton 6 described Zn(s) reduction the at the Zn"+(aq) following cell cathode; Cd?+(aq) notation: 3content artribucion
Provide Write 2 the Question Content attribution L your answer overall Interpret Cell Notation to Describe _ (aq ) below: reaction that ~Caq) Rencton 6 described Zn(s) reduction the at the Zn"+(aq) following cell cathode; Cd?+(aq) notation: 3 content artribucion


Answers
Write a balanced equation for the overall cell reaction, and give a brief description of a galvanic cell represented by the following shorthand notation:
$$\mathrm{Pb}(s)\left|\mathrm{Pb}^{2+}(a q)\right|\left|\mathrm{Br}_{2}(l)\right| \mathrm{Br}^{-}(a q) | \mathrm{Pt}(s)$$
Okay, So for this problem, we're going to calculate the sander's self potential, right? The overall balance cell reaction. So the first step is to go to table 20.1 in your textbook and look up the values for these reduction potentials for the half reactions. So you'll see for the reaction with platinum. Yeah. Have you? Value is one point one age and the reaction of 10. You have negative 0.1 three 75 volts. Eso the half reaction with lower reduction potential will proceed in the opposite direction. It will be the oxidation reaction. So, as you see, negative 0.1 375 is less than 1.18 So this one, the 10 reaction is going to proceed in the opposite direction. So she draw the to write the overall reaction. First thing we need to do is, um, reverse the tin reactions that we're gonna have. Yeah, solid tune on this side on, then we can add that to the platinum reaction is gonna proceed as written. Okay. Yeah. Okay. So the electrons, we're gonna cancel, and we can rewrite this as the final answer. Mhm. No. All right. Mhm. Yeah, Yeah. Yeah. Okay, So this is the overall cell reaction. Now, to write this using sell notation, you find the the reaction that took place of the node goes on the left, which is the accid ation reaction. She start with tin solid. Yeah, you dropped one parallel line to represent the phase boundary the difference between the two phases and then mhm two parallel lines. Then we write the reaction that occurs at the CAFO, which is the reduction. Mhm another wine. And then you write the last part right here. Mhm. Now, to calculate the standard self potential. Yeah, just, um, what you're gonna do is this is e equal. Yeah. Uh huh. To the reduction minus No. Um, so you need the standard self if you look at the values we got from the chart, so we would use the value that goes with the reduction reaction, which was 1.18 and then you're going to subtract it from the value that we saw for the oxidation zero. Thanks. Mhm. Now, to my choices are going to cancel each other so you end up adding them together. And you she at one point 316 volts
The question here asks us to write a balanced cell reaction but the sale notation shown here and to calculate the standard self potential for this and then using the self potential to deter mine if this is going to be a spontaneous reaction or a non spontaneous reaction under standards, state conditions. So if we start with writing the cell reaction way, have to remember that the same notation indicates that the reaction takes place from left to right. So let's take the left most copper species and the left most gold species and put them on the left hand side of the reaction. And that will give us this Papa solid. Thus good three plus equals. I mean, like, there's a little smaller. So this enough room toe balance the equation and this goes to company two. Plus, it's just a right most copper species. Hey, Chris and the right most whole species. Good, solid. And this is a huge Okay, now let's start to balance this equation. So we see that the charges, um oh, three plus on this side and two plus on the right hand side. So if we started with balancing the charges, we need to have the same number of charges on the two sides, and we can do that by adding to in front of go three plus and three in front of copper two plus. So now both sides have six plus as the charge. Now we can balance the number of atoms, so we see that there's tool gold species on the left hand side. So let's put it to in front of the, um, good solid. And then we see that there's three copper species on the right hand side, and so we'll have to add three in front of the copper solid. And this is the balanced equation. So once we have this, um, because we know that the hot dog is the salutation, which is given on the right hand side on the A node is the self notation part that comes on the left hand side. We can use that information to calculate the self potential so the self potential itself is going to be equal to do potential off the catalog. When I kept for shot and potential after and what has to be subjected by that and because we know that the gold species is the tattooed on the, um, copper species. From the annoyed, we can use their reduction potentials to calculate the self potential. The reduction potential off the gold catalog is going to be one blind food 98 Once this is something we can look up for, the reaction off Good three plus getting converted to a gold solid and the honored reduction potential is positive. Find 34 words. So we put the subject subtraction sign from here. That's why there's a negative sign in front of 0.3 food. But other than that 0.34 itself is a positive value. Get so when we take the difference of these two, we get one. Find one six watch. This is the self potential, and it's a positive value because it's a positive value. It indicates that this is a spontaneous reaction.
Okay, So in this video, we're going to be answering question two from chapter 20 which asked us to for each pair of half reactions right the balanced equation for the overall cell reaction and calculate the standard cell potential. So the two reactions were given our seal plus two plus two electrons forming neutral CEO and seer plus three close three electrons forming CR um, and we can look up the cell potentials for those that they're negative 0.28 and negative 0.744 respectively. Um, and basically, we're going to manipulate these 2/2 reactions and then add them together. The reason we're going to manipulate the first is that we want the number of electrons to cancel. So we want the same number of electrons to appear on both sides of the equation and for those to cancel out. Um, so as we manipulate the half reactions, we want to keep track of what's happening. Thio Ourself potential. Um, so if we multiply 1/2 reaction by a number, there's no change in the cell potential. We leave. It just is it is that reduction potential on def. We flipped the direction of 1/2 reaction. Then we flip the sign on the reduction potential. Um and so we were given to reduction reaction. So for this toe, add together to form a Redox reaction, we're going to have to flip the direction on one of them. So we need toe pick. Which one is going to be our oxidation reaction and change directions on dhe? To do that, we generally as a convention we picked the half reaction with lowest reduction potential on dhe. That's the one that's going to flip direction, toe oxidation, and that will give us the the highest possible you not the most positive you not on. So remember, we want the number of electrons to cancel. So because we had two electrons in the first half reaction in three electrons in the second half reaction, we're gonna multiply the first half reaction by three in the second half, reaction by two. And then remember, nothing's goingto happen to our reduction potentials. As a result of that, they're going to stay exactly the same. So now we have to have reactions with the same number of electrons, which is great, but in order for those same number of electrons to cancel out. They need to appear on opposite sides of the equation. We have two reduction equations to reduction Have reactions where we're gaining electrons on and we need one of those to be a loss of electrons. We need one of those to be oxidation to be going in the reverse direction. So we're gonna pick the line with the lowest reduction potential to change sign arm. That'll give us the most positive self potential. So we're picking the lowest reduction potential, and that's the one that's gonna flip. Um, so then we're left with three CEO plus two plus six electrons forming three CEO and to see are being oxidized, losing six electrons to form to see our plus three. And then we change the sign on that reduction potential s So then when we add them together, we get a cell potential of 0.464 volts, um, and are electrons are six electrons that appear on both sides of our equation? Cancel out and were left with three CEO plus two plus two c R forms. Three CEO plus two c R plus three, which is balanced
If all we have is copper, two plus and copper solid and still get electrons to flow and what we have, what's called a concentration. So so we could have see you two plus being reduced to see you solid at the cathode. And we could have seen solid being oxidized to see you two plus. But theano the net reaction then would look something like this with a reactant in the products of the same thing.